## Precalculus (6th Edition) Blitzer

Let us consider the given expression on the left side $\sin t-\cos 2t$, which can be simplified by using the double angle formula $\cos 2t=1-2{{\sin }^{2}}t$. \begin{align} & \sin t-\cos 2t=\sin t-\left( 1-2{{\sin }^{2}}t \right) \\ & =\sin t-1+2{{\sin }^{2}}t \\ & =2{{\sin }^{2}}t+\sin t-1 \\ & =\left( 2\sin t-1 \right)\left( \sin t+1 \right) \end{align} Hence, the expression on the left-side is equal to the expression on the right-side.